Answer:
3.8s
Step-by-step explanation:
Solve for h=0, you will get t=3.75
A standard number cube has 6 sides, and so the numbers are:
1,2,3,4,5,6
P, is any of those numbers that <em>are not 2</em>.
2 is one of the numbers in 6, and so it is 1/6 of the die.
6/6 - 1/6 = 5/6
C) 5/6 is your chance of getting a number that is not 2
hope this helps
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
Where is X in the equation?
